Determine all possible values of a prime number P such that P
3-4P+9 is a perfect square.
Note: Computer program aided solutions are welcome, but a semi-analytic solution is preferred.
We have P^3-4P+9 = N^2. P and N are both positive integers and P is prime.
Subtract 9 from each side and factor:
(P-2)*P*(P+2) = (N-3)*(N+3)
P is prime therefore it must divide N-3 or N+3. Then we have two similar cases: N-3=P*F or N+3=P*G. Substitute back and the equation then becomes
(P-2)*(P+2) = F*(P*F-6) or (P-2)*(P+2) = G*(P*G+6)
Next, lets complete the square on P for both equations:
(2P-F^2)^2 = F^4-24F+16 or (2P-G^2)^2 = G^4+24G+16
Lets look at the first equation (2P-F^2)^2 = F^4-24F+16. F is a positive integer, so the right side F^4-24F+16 < F^4. We also need the right side to be a square which then must be at most (F^2-1)^2.
Therefore F^4-24F+16 <= (F^2-1)^2.
Solving the inequality gives us a range of F to be -15.35 <= F <= 27.35. But F is a positive integer, so only F = 1,2,...,27 are possible. Trying each of these in F^4-24F+16 gives exactly one F that results in a perfect square: F=3. Then (2P-3^2)^2 = 5^2, which yields P=2 or 7.
Now we can do a similar thing to (2P-G^2)^2 = G^4+24G+16. In this case we conclude (G^2+1)^2 <= G^4+24G+16.
This inequality solves to -20.38 < G < 32.38. Again, G is a positive integer, so only G = 1,2,...,32 are possible. And trying each of these in G^4+24G+16 yields a square for exactly one value when G=3. Then (2P-G^2)^2 = 13^2, which yields P=-2 or 11.
So now we are down to four potential values of P: -2, 2, 7, 11. -2 must be discarded as P must be positive. Each of the other values checks out:
P=2: 2^3-4*2+9 = 9 = 3^2
P=7: 7^3-4*7+9 = 324 = 18^2
P=11: 11^3-4*11+9 = 1296 = 36^2
Edit: Fixed some typos.
Edited on December 12, 2022, 10:46 am
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